Suppose $\sum a_n$ converges absolutely and $\sum b_n$ converges. Give an example where the Cauchy product does not converge absolutely.

Problem:

Suppose $\sum_{n=0}^{\infty} a_n$ converges absolutely and $\sum_{n=0}^{\infty} b_n$ converges. Give an example where the Cauchy product, defined as $\sum_{n=0}^{\infty} c_n$ where $c_n=\sum_{k=0}^{n} a_k b_{n-k}$, does not converge absolutely.

Question:

I have been scratching my head over this for a while. What would an example be?

• @TedShifrin That doesn't answer the question; this is about showing that if only one of the two series are absolutely convergent, then it is possible for the Cauchy product to converge only conditionally. – Hayden May 9 '17 at 2:22
• @Hayden: Doesn't the question ask for an example where the Cauchy product "does not converge absolutely"? – Ted Shifrin May 9 '17 at 5:05
• The suggested duplicate addresses a different question. There the counterexample uses series neither of which is absolutely convergent. – RRL May 9 '17 at 6:48
• @TedShifrin Yes, and while "divergent" implies "does not converge absolutely", Merten's Thm says that the OP's conditions are enough for convergence, so the "duplicate" doesn't fit the bill. – Hayden May 9 '17 at 14:05
• I've reopened this question since it seems quite different than the cited duplicate. – robjohn Mar 2 '18 at 13:35

Take $a_n = (-1)^n/n^2$ and $b_n = (-1)^n/n$. Then $\sum a_n$ converges absolutely and $\sum b_n$ converges conditionally.

The Cauchy product

$$\sum_{n=1}^\infty \sum_{k=1}^n \frac{(-1)^k}{k^2} \frac{(-1)^{n+1-k}}{n+1-k} = \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=1}^n \frac{1}{k^2(n+1-k)},$$

converges by the AST since the terms $\sum_{k=1}^n 1/(k^2(n+1-k))$ are decreasing.

We also know the Cauchy product converges by the general theorem guaranteeing convergence if one series is absolutely convergent and the other is convergent.

However, the Cauchy product is not absolutely convergent since

$$\sum_{n=1}^\infty\sum_{k=1}^n \frac{1}{k^2(n+1-k)} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{ 1}{3} + \frac{1}{8} + \frac{1}{9} + \ldots > 1 + \frac{1}{2}+ \frac{1}{3} + \ldots$$

diverges.

• How do we know that $\lim_{n \to \infty}\sum_{k=1}^{n}\frac{1}{k^2(n+1-k)} \to 0$? I have tried comparison test but can't find anything nice to compare against. – EternusVia May 10 '17 at 0:15
• It has to because the Cauchy product series converges. One series is absolutely convergent, the other is convergent and by Merten's theorem, the Cauchy product converges. Showing absolute divergence is simply a consequence of $\sum_{n=1}^\infty \sum_{k=1}^n 1/(k^2(n+1-k) > \sum_{n=1}^\infty \sum_{k=1}^1 1/(k^2(n+1-k) > \sum_{n=1}^\infty 1/n = \infty$. So you have the example you requested. – RRL May 10 '17 at 1:07
• That makes sense!! Thanks. – EternusVia May 10 '17 at 3:07
• Also one could divide the sum: when $k\le \frac{n+1}2$ you have $n+1-k\ge \frac{n+1}2$ and hence $\sum_{k=1}^{(n+1)/2} \frac1{k^2(n+1-k)} \le \frac2{n+1} \sum_{k=1}^{(n+1)/2} \frac1{k^2} \le \frac2{n+1} \zeta(2)$. The other part is easier: $n+1-k\ge 1$ and $\sum_{k=(n+1)/2}^{n}\frac1{k^2(n+1-k)} \le \sum_{k=(n+1)/2}^{n}\frac1{k^2} \le \sum_{k=(n+1)/2}^{\infty}\frac1{k^2} \sim \frac2 n$ (a rest of a convergent series). Actually this quantity is equivalent to some constant over $n$ but the upperbound is enough. – Emmanuel Amiot Jan 2 '18 at 10:28
• Sorry, I am wondering how to show the terms $\sum_{k=1}^n 1/(k^2(n+1-k))$ are decreasing? – Tony Ma Jun 18 '18 at 13:57

Hint: Consider the Cauchy product of $1+0+0+0+\dots$ and $1-\frac12+\frac13-\frac14+\dots$.

We can take this even further: $$\overbrace{\sum_{k=0}^\infty\frac{1+(-1)^k}{2^{k+1}}}^{\substack{\text{converges}\\\text{absolutely}}}\,\,\overbrace{\sum_{j=0}^\infty\frac{(-1)^j}{j+1}}^{\substack{\text{converges}\\\text{conditionally}}}$$ The Cauchy product is $$\sum_{j=0}^\infty(-1)^j\overbrace{\sum_{k=0}^{\lfloor j/2\rfloor}\frac1{j-2k+1}\frac1{2^{2k}}}^{a_j\ge\frac1{j+1}}$$ where, for $j\ge5$, \begin{align} a_{j-1}-a_j &=\sum_{k=0}^{\lfloor(j-1)/2\rfloor}\frac1{j-2k}\frac1{2^{2k}}-\sum_{k=0}^{\lfloor j/2\rfloor}\frac1{j-2k+1}\frac1{2^{2k}}\\ &=\sum_{k=0}^{\lfloor(j-1)/2\rfloor}\frac1{(j-2k)(j-2k+1)}\frac1{2^{2k}}-\frac{1+(-1)^j}{2^{j+1}}\\ &\ge\frac1{j(j+1)}-\frac1{2^j}\\[12pt] &\ge0 \end{align} Thus, the Cauchy product converges by the Dirichlet Test, but does not converge absolutely since $a_j\ge\frac1{j+1}$